Connectedness in Metric Spaces

Metric Spaces

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Connectedness

Learn the concept of Connectedness in Metric Spaces through formal definitions, key theorems, and equivalent characterizations with proofs.

Disconnected Metric Space

Let \( (\mathbf{X}, d ) \) be a metric space. Then \( \mathbf{X} \) is said to be a disconnected metric space if there exist two non-empty subsets \( A \) and \( B \) such that:

\( \mathbf{X} = A \cup B \)
\( A \cap \overline{B} = \Phi \) and \( \overline{A} \cap B = \Phi \)

Connected Metric Space

Let \( (\mathbf{X}, d ) \) be a metric space. Then \( \mathbf{X} \) is said to be a connected metric space if it is not a disconnected metric space.

Disconnected Metric Space Implies Union of Two Disjoint Open Sets

Let \( (\mathbf{X}, d ) \) be a metric space. If \( \mathbf{X} \) is disconnected, then there exist two non-empty disjoint open subsets \( A \) and \( B \) such that:

\mathbf{X} = A \cup B

Given that \( (\mathbf{X}, d ) \) is a metric space and is disconnected.

Then there exist two non-empty subsets \( A \) and \( B \) such that:

\( \mathbf{X} = A \cup B \)
\( A \cap \overline{B} = \Phi \) and \( \overline{A} \cap B = \Phi \)

To prove: \( A \) and \( B \) are open.

\begin{align*} & \mathbf{X} = A \cup B \\ \Rightarrow\ & \mathbf{X} - \overline{B} = (A \cup B) - \overline{B} \\ \Rightarrow\ & \mathbf{X} - \overline{B} = (A - \overline{B}) \cup (B - \overline{B}) \\ \Rightarrow\ & \mathbf{X} - \overline{B} = A \cup \Phi \quad \text{since } A \cap \overline{B} = \Phi \text{ and } B \subseteq \overline{B} \\ \Rightarrow\ & \mathbf{X} - \overline{B} = A \end{align*}

Since \( \overline{B} \) is closed, \( \mathbf{X} - \overline{B} \) is open. Therefore, \( A \) is open.

\begin{align*} & \mathbf{X} = A \cup B \\ \Rightarrow\ & \mathbf{X} - \overline{A} = (A \cup B) - \overline{A} \\ \Rightarrow\ & \mathbf{X} - \overline{A} = (A - \overline{A}) \cup (B - \overline{A}) \\ \Rightarrow\ & \mathbf{X} - \overline{A} = \Phi \cup B \quad \text{since } \overline{A} \cap B = \Phi \text{ and } A \subseteq \overline{A} \\ \Rightarrow\ & \mathbf{X} - \overline{A} = B \end{align*}

Since \( \overline{A} \) is closed, \( \mathbf{X} - \overline{A} \) is open. Therefore, \( B \) is open.

Open Disconnection Implies Closed Disconnection

Let \( (\mathbf{X}, d ) \) be a metric space. If there exist two non-empty disjoint open subsets \( A \) and \( B \) such that:

\mathbf{X} = A \cup B

then there exist two non-empty disjoint closed subsets \( C \) and \( D \) such that:

\mathbf{X} = C \cup D

Given that \( (\mathbf{X}, d ) \) is a metric space and there exist two non-empty disjoint open subsets \( A \) and \( B \) such that \( \mathbf{X} = A \cup B \).

To prove: \( A \) and \( B \) are also closed.

Since \( A \cap B = \Phi \), we have:

\( A = \mathbf{X} - B \) and \( B = \mathbf{X} - A \).

Since \( A \) and \( B \) are open, their complements \( \mathbf{X} - A \) and \( \mathbf{X} - B \) are closed.

Hence, \( A \) and \( B \) are both closed.

Disjoint Decomposition Implies Existence of Clopen Set

Let \( (\mathbf{X}, d ) \) be a metric space. If there exist two non-empty disjoint subsets \( A \) and \( B \) such that:

\mathbf{X} = A \cup B

then there exists a proper subset of \( \mathbf{X} \) which is both open and closed in \( \mathbf{X} \).

Given that \( (\mathbf{X}, d ) \) is a metric space and there exist two non-empty disjoint closed subsets \( A \) and \( B \) such that \( \mathbf{X} = A \cup B \).

To prove: \( A \) and \( B \) are both open and closed in \( \mathbf{X} \).

Since \( A \cap B = \Phi \), we have:

\( A = \mathbf{X} - B \) and \( B = \mathbf{X} - A \).

Since \( A \) and \( B \) are closed, their complements \( \mathbf{X} - A \) and \( \mathbf{X} - B \) are open.

Hence, \( A \) and \( B \) are both open and closed (clopen) in \( \mathbf{X} \).

Existence of Clopen Set Implies Disconnectedness

Let \( (\mathbf{X}, d ) \) be a metric space. If there exists a proper subset of \( \mathbf{X} \) which is both open and closed in \( \mathbf{X} \), then \( (\mathbf{X}, d ) \) is disconnected.

Given that \( (\mathbf{X}, d ) \) is a metric space and let \( A \) be a proper subset of \( \mathbf{X} \) which is both open and closed.

Let \( B = \mathbf{X} - A \). Then:

\( \mathbf{X} = A \cup B \)
\( A \cap B = \Phi \)

To prove: \( (\mathbf{X}, d ) \) is disconnected.

Since \( A \) is closed, \( A = \overline{A} \), so \( \overline{A} \cap B = \Phi \).

Since \( A \) is open, \( B \) is closed, so \( B = \overline{B} \), hence \( A \cap \overline{B} = \Phi \).

Therefore, \( \mathbf{X} = A \cup B \) with \( \overline{A} \cap B = \Phi \) and \( A \cap \overline{B} = \Phi \).

Thus, \( (\mathbf{X}, d ) \) is disconnected.